Sign up ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.


Limit[2^x Log[x], x -> 0, Direction -> -1] 

(right limit) gives correctly the answer


but the command

Limit[2^x Log[x], x -> 0, Direction -> 1] 

(left limit) gives again


The same in WolframAlpha. What is the reason behind this ?

share|improve this question

3 Answers 3

You are correct in that the limit of the complex logarithm on the standard branch, as $x\to 0$ from below, should better be written as $-\infty + i \pi$.

The reason why Mathematica doesn't give this result is that it uses a polar representation for infinity in the complex plane. You can see this by doing

FullForm[Limit[2^x Log[x], x -> 0, Direction -> 1]]


This DirectedInfinity expresses the fact that the imaginary part goes to a constant whereas the real part diverges.

However, in complex analysis it is often necessary to retain constants such as the one in $-\infty + i \pi$, e.g., to choose integration paths correctly. If you want to get that constant from a Limit, I think the easiest way to do it is to forcibly do the limits of the real and imaginary parts separately:

 Limit[h[2^x Log[x]], x -> 0, Direction -> 1], {h, {Re, Im}}]

(* ==> {-Infinity, Pi} *)
share|improve this answer

It's a correct result.It is best to illustrate on the plot.

enter image description here

f[x_] := 2^x*Log[x];
{Limit[f[x], x -> 0, Direction -> 1], 
Limit[f[x], x -> 0, Direction -> -1]}

$\{-\infty ,-\infty \}$

Plot[{Re@f[x], f[x]}, {x, -2, 2}, Filling -> Bottom, 
PlotRange -> {{-2, 2}, {-10, 3}}, PlotLegends -> "Expressions"]
share|improve this answer

It's a correct result. The computation below may help to see this.

Table[2^x Log[x], {x, -(10.^Range[-1, -10, -1])}]
(* Out[48]= {-2.14838785758 + 2.93120959177 I, -4.57335995192 + 
  3.119892088 I, -6.90296884693 + 3.13941582202 I, -9.20970198196 + 
  3.14137490253 I, -11.5128456637 + 3.1415708778 I, -13.8155009818 + 
  3.141590476 I, -16.1180945337 + 3.14159243583 I, -18.4206806163 + 
  3.14159263181 I, -20.7232658226 + 3.14159265141 I, -23.0258509283 + 
  3.14159265337 I} *)

The values are heading to negative infinity, on a path near Pi units north of the negative axis. Remark: for better or worse, Mathematica does not have a notion of "infinite real part plus finite imaginary part", so -Infinity+I*Pi just becomes -Infinity.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.